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A have a weighted die and I toss it $N$ times. The empirical probability of getting the $i^{\text{th}}$ face of the dice is: $$ P(i) = \frac{N_i}{N},$$ where $N_i$ is the number of times I tossed $i$ and $N$.

Now, if I think of the empirical probability as an estimate of the "true probability" (whatever that might mean) - how can I estimate its $variance$? I've looked everywhere, but to no avails.

In particular, I expect that such a procedure would take into account that $\sum_{i=1}^6 P(i)=1$.

EDIT

I apologize I didn't make that clear:

I repeat the procedure $M$ times: toss the dice $N$ times, calculate the emprirical probability $P_m(i)$ (for each procedure $m$). The variance I was trying to ask about is the variance in the empirical probabilities $\{ P_m(i)\}$, let's call them $\sigma^2_m(i)$.

Is there a way of estimating or calculating that from a single trial?

ponadto
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1 Answers1

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Denote the true probability by $p_{i}$ and note that $N_{i}$ is binomially distributed with parameters $p_{i}$ and $N$. Its expectation is $Np_{i}$ and its variance is $Np_{i}\left(1-p_{i}\right)$. For $P_{i}=\frac{N_{i}}{N}$ you find: $E\left[P_{i}\right]=\frac{EN_{i}}{N}=\frac{Np_{i}}{N}=p_{i}$ showing that it is an unbiased estimator of $p_{i}$. Secondly: $\text{Var}\left(P_{i}\right)=\frac{\text{Var}\left(N_{i}\right)}{N^{2}}=\frac{Np_{i}\left(1-p_{i}\right)}{N^{2}}=\frac{p_{i}\left(1-p_{i}\right)}{N}$.

Here you are not estimating but determining a variance, so I am not sure wether this is what you are looking for.

drhab
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